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21    PRODUCT OF MEASURES       227

with the product convention referred to above. Hence it is enough to prove the



Now this inequality is clearly valid when the right-hand side is equal to -f oo.
So consider the case in which each of the factors on the right is finite.
Then there exist two decreasing sequences (/), (#) such that fn e ./(X),
)9f^fH9ff^gH for all n, and


r*                 r*                r*                   /*

/Ai=lim    /ndA,           #^ = lim

J                      n~xx> J                              J                       />-+oo J

By reason of our conventions and of (12.7.5), the function (x, y)*~~*fn(x)gn(y)
belongs to */(X x Y), and we have f(x)g(y) ^fn(x)gn(y) for all n and all
(x9 j) e X x Y. But, by virtue of (13.21.3),

f *


- ff*
JJ "X9

from which the desired inequality follows by letting n tend to +00.

Finally, to deal with the case where (for example) / is ^-negligible, it
is enough (by virtue of our conventions) to prove that the function
(x,y)\-+f(x)g(y) is v-negligible. This is a consequence of the following

(13.21.12)    If N is any ^-negligible subset of X, then the set N x Y is v-

For since Y is the union of a denumerable sequence (Ln) of compact sets,
it is enough to show that each of the sets N x Ln is v-negligible. But since we
have proved that ( is valid whenever both of the factors on the
right-hand side are finite, we may take/= <pN and g = <pLn in this formula.

(13.21.13)    Let E, F, G be three topological spaces and u : E x F - G a con-
tinuous mapping. L#//(resp. g) be a ^-measurable mapping ofX into E (resp. a
^-measurable mapping ofY into F). Then (x, y)^u(f(x),g(yj) is a v-measur-
able mapping ofX x Y into G.                                     '                        J